Solução_Calculo_Stewart_6e

F.
216 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
15. True. Let x 1,x 2 ∈ I and x 1 1
f(x 2 )
[f is positive] ⇒
17. True. If f is periodic, then there is a number p such that f(x + p) =f(p) for all x. Differentiating gives
f 0 (x) =f 0 (x + p) · (x + p) 0 = f 0 (x + p) · 1=f 0 (x + p),sof 0 is periodic.
19. True. By the Mean Value Theorem, there exists a number c in (0, 1) such that f(1) − f(0) = f 0 (c)(1 − 0) = f 0 (c).
Since f 0 (c) is nonzero, f(1) − f(0) 6= 0,sof(1) 6=f(0).
1. f(x) =x 3 − 6x 2 +9x +1, [2, 4]. f 0 (x) =3x 2 − 12x +9=3(x 2 − 4x +3)=3(x − 1)(x − 3). f 0 (x) =0 ⇒
x =1or x =3,but1 is not in the interval. f 0 (x) > 0 for 3